HMMT 十一月 2024 · 团队赛 · 第 8 题

HMMT November 2024 — Team Round — Problem 8

[50] Compute the unique real number x < 3 such that √ √ √ (3 − x )(4 − x ) + (4 − x )(6 − x ) + (6 − x )(3 − x ) = x.

解析

[50] Compute the unique real number x < 3 such that p p p (3 − x )(4 − x ) + (4 − x )(6 − x ) + (6 − x )(3 − x ) = x. Proposed by: Pitchayut Saengrungkongka Answer: 23 / 8 = 2 . 875 √ √ √ Solution 1: Let a = 3 − x , b = 4 − x , c = 6 − x , so x = ab + bc + ca . Then 2 ( a + b )( a + c ) = a + ab + bc + ca = (3 − x ) + x = 3 . Likewise, we get that ( a + b )( a + c ) = 3 ( b + a )( b + c ) = 4 ( c + a )( c + b ) = 6 . By multiplying these equations and taking the square root, we get that √ √ ( a + b )( b + c )( c + a ) = 72 = 6 2 , so √ √ √ 3 2 b + c = 2 2 , c + a = , a + b = 2 , 2 and hence √ √ √ √ 3 2 2 2 a = + 2 − 2 2 = ⇒ a = . 2 4 √ 23 Since a = 3 − x , it follows that x = . 8 Solution 2: The given equation implies √ √ √ 2 3 − x + 4 − x + 6 − x = (3 − x ) + (4 − x ) + (6 − x ) p p p

2 (3 − x )(4 − x ) + (4 − x )(6 − x ) + (6 − x )(3 − x ) = 13 − 3 x + 2 x = 13 − x. √ √ √ As 3 − x + 4 − x + 6 − x is nonnegative, √ √ √ √ 3 − x + 4 − x + 6 − x = 13 − x. We repeatedly rearrange, square both sides, and simplify: √ √ √ √ 3 − x + 4 − x = 13 − x − 6 − x p p 7 − 2 x + 2 (3 − x )(4 − x ) = 19 − 2 x − 2 (13 − x )(6 − x ) p p (3 − x )(4 − x ) = 6 − (13 − x )(6 − x ) p (3 − x )(4 − x ) = 36 − 12 (13 − x )(6 − x ) + (13 − x )(6 − x ) p 2 x − 17 = 2 (13 − x )(6 − x ) 2 4 x − 68 x + 289 = 4(13 − x )(6 − x ) 23 8 x = 23 = ⇒ x = . 8